一元二次方程x^2-2ax+a+6=0的两实根为x1,x2,求函数f(a)=(x1-1)^2+(x2-1)^2的取值范围

判别式=4a^2-4(a+6)>=0
a^2-a-6>=0
(a-3)(a+2)>=0
a<=-2或a>=3
x1+x2=2a,x1*x2=a+6
x1^2+x2^2=(x1+x2)^2-2x1*x2=4a^2-2a-12
f（a）=（x1-1）^2+(x2-1)^2
=(x1^2+x2^2)-2(x1+x2)+2
=4a^2-2a-12-4a+2
=4a^2-6a-10
=4(a-3/4)^2-49/4
因为a<=-2或a>=3,
(a-3/4)^2>=(9/4)^2
4(a-3/4)^2-49/4>=8
f（a)>=8